which of the statements is true?\noa. as x approaches positive infinity, f(x) exceeds g(x) and h(x).\nob. as…

which of the statements is true?\noa. as x approaches positive infinity, f(x) exceeds g(x) and h(x).\nob. as x approaches positive infinity, g(x) exceeds f(x) and h(x).\noc. as x approaches positive infinity, h(x) converges with g(x).\nod. as x approaches positive infinity, h(x) exceeds f(x) and g(x).

which of the statements is true?\noa. as x approaches positive infinity, f(x) exceeds g(x) and h(x).\nob. as x approaches positive infinity, g(x) exceeds f(x) and h(x).\noc. as x approaches positive infinity, h(x) converges with g(x).\nod. as x approaches positive infinity, h(x) exceeds f(x) and g(x).

Answer

Explanation:

Step1: Analyze the growth rate of linear, exponential and logarithmic functions

  • (f(x)) is a linear function ((y = x - 1)). The general form of a linear function is (y=mx + b) with a constant slope (m). As (x\to+\infty), (y = mx + b) grows at a constant rate.
  • (g(x)) is an exponential - like function (assuming (g(x)=2^{x}) based on its shape). The general form of an exponential function is (y = a\cdot b^{x}+c) ((b> 1)). As (x\to+\infty), (y=a\cdot b^{x}+c) ((b > 1)) grows much faster than a linear function. The growth rate of (y=a\cdot b^{x}+c) ((b>1)) is proportional to the function's current value.
  • (h(x)) is a logarithmic - like function (assuming (h(x)=\log_{2}(x + 3)) based on its shape). The general form of a logarithmic function is (y=\log_{b}(x - h)+k). As (x\to+\infty), (y = \log_{b}(x - h)+k) grows, but its growth rate (\frac{dy}{dx}=\frac{1}{(x - h)\ln b}) approaches (0).

Step2: Compare the functions as (x\to+\infty)

  • For a linear function (y_1=mx + b) ((m>0)), an exponential function (y_2=a\cdot b^{x}+c) ((b > 1,a>0)) and a logarithmic function (y_3=\log_{d}(x - k)+l) ((d>1)):
    • (\lim_{x\to+\infty}\frac{y_1}{y_2}=\lim_{x\to+\infty}\frac{mx + b}{a\cdot b^{x}+c}). Using L'Hopital's rule (for (\frac{\infty}{\infty}) form, if we consider the ratio of the derivatives), (\lim_{x\to+\infty}\frac{m}{a\cdot b^{x}\ln b}=0). So (y_2) (exponential - type function (g(x))) grows faster than (y_1) (linear function (f(x))).
    • (\lim_{x\to+\infty}\frac{y_3}{y_2}=\lim_{x\to+\infty}\frac{\log_{d}(x - k)+l}{a\cdot b^{x}+c}). Since the derivative of (\log_{d}(x - k)+l) is (\frac{1}{(x - k)\ln d}) and the derivative of (a\cdot b^{x}+c) is (a\cdot b^{x}\ln b), (\lim_{x\to+\infty}\frac{\frac{1}{(x - k)\ln d}}{a\cdot b^{x}\ln b}=0). So (y_2) (exponential - type function (g(x))) grows faster than (y_3) (logarithmic - type function (h(x))).

Answer:

B. As (x) approaches positive infinity, (g(x)) exceeds (f(x)) and (h(x))